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CURRENT OBJECTIVE Use the properties of the definite integralQuestion15 15 f(t) dt given that f(t) dt = -10 and f(t) dt = 9.Determine the value ofProvide your answe...

Question

CURRENT OBJECTIVE Use the properties of the definite integralQuestion15 15 f(t) dt given that f(t) dt = -10 and f(t) dt = 9.Determine the value ofProvide your answer below:

CURRENT OBJECTIVE Use the properties of the definite integral Question 15 15 f(t) dt given that f(t) dt = -10 and f(t) dt = 9. Determine the value of Provide your answer below:



Answers

Evaluate the definite integral. Use a graphing utility to verify your result. $$\int_{0}^{3}\left(t-5^{t}\right) d t$$

In this problem. They want us to evaluate the integral of Tante to the Fifth Ministry whole thing divided by T DT. So here it should be pretty clear that we probably need to divide each term in the numerator by this denominator of tea. So this gives us the integral of well, pin to the 10 t to the fifth, divided by teas just 10 t to the fourth. And I'm going to go ahead and split this into two Integral so D t minus the Negro of three over tea DT So here there's nothing to counsel, so we can just leave it like that, right? We can also write the tea as a teacher of the minus one exponents will to play in the three. But let's just leave it like that. So the tennis is Ah factor, some Confederate out of the integral and see to the fourth is a polynomial, so we can just apply the power role here. So we add one to the exponents and we divide by the new exponents, which is five and then here we have minus that three is a constant, so that can come out. Constant factor that is on Dean one over T. The anti derivative of one over tea is the natural log of the absolute value of tea, plus the arbitrary constant of integration. So now we just divide him by five. So that gives us two so to t to the fifth minus three times the natural log of the absolute value of tea plus e. And that's the answer.

It's a problem this you wanna write. It's an integral integral from their own to t I. A minus is plus three inches off. So the definitions assistant coach is integral to I mean, come on, your business. It's a girl from zero to two. Once integral Teo too. He's killed you. Steer us integral from there to to drink his Okay. Me too. Half He's acquired from zero to I once force in truth for from zero to to see us one we're going to just trust thinks they're all too true. House came in this country Sure, Linus or plus thirty two.

Find us integral here will need to use integration by parts. So we're going to let you be equal to t squared & DVE five T. Okay, so D you is to T D. T And we is 1/5 each of the fight. So now putting this into our integration by parts formula. This will give us 1/5. He squared Times each of the five T minus the integral of We do. So let's put the constants for the front. So that would be 2/5 integral of E to the five teeth. Okay, so copy down B first part -2-5. And to find that integral will have to use integration by parts again. So this time we're letting you be equal to T. So that means d'you is D T do we is E to the five td T. So be is again 1/585. So now let's put that into the integration by parts formula again. So this will be U. Times V. So that's 1/5. The times we need to the five p minus the integral of V. D. You. So let's pull the 1/5 in front of the integral. And we have the integral of E. For the five T. V. Let's go ahead and simplify this a bit. This is 1/5 T squared. Each of the five teeth, Distribute the negative to over five and we get to over 25 T. E. To the fight T plus two over 25. And the integral of each of the five T. Is just 1/5. Need a fighting and then we have to add a constant C. So therefore our final answer is 1/5. He squared each of the five T minus when to over 25 T. Times E. To the five T. And then plus two over 125. Me Too, the five T plus

Definite integral oh T to the power five. Now you do de que plus one DT from 1 to 2. So here we have to evaluate the definite intrigue. Als. So here the fundamental theorem of calculus is definite integral of F. Of X. Dx from A to B equal to F O V 90 F of a will capital F is anti derivative of F. So and and did anybody off T to the power of five. Now I do dig you plus one is two to the power six upon six negative two to the power four upon four plus the therefore using the fundamental theorem of calculus. So here we get definite integral of due to the about five negative D. You plus one DT from 1 to 2 equal to due to the power six upon six negative due to the bar for upon four plus T from 1 to 2. So first of all we plug in the upper value which is equal to two than we get two to the power six upon six. Now you do due to the power four upon four plus. Do no, I do. Went to the power six upon six, negative one. To the power four upon four plus one. Now will simplify this and here we get 26 upon three. Now you do 11.12 equal to 31 upon four. So it is over. Final answer.


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